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3 years ago

What is the specific heat of the solid phase? (Please see picture attached)

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

B.0.2 J/g°C

Explanation:

From the attached picture;

Heat attained in the solid phase is 200 Joules
Change in temperature is 50°C ( from 0°C to 50°C)
Mass of the solid is 20 g

We are required to determine the specific heat capacity of the substance;

We need to know that Quantity of heat is given by the product of mass,specific heat capacity and change in temperature.
That is; Q = mcΔT

Rearranging the formula;

c = Q ÷ mΔT

Therefore;

Specific heat = 200 J ÷ (20 g × 50°c)

= 0.2 J/g°C

Thus, the specific heat of the solid is 0.2 J/g°C

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determine mass of water formed when 12.5 L NH3(at298K and 1.50atm) is reacted with 18.9L of O2 (at 323K and 1.1atm)

sasho [114]

The mass of water formed is

calculation

Use the ideal gas equation to calculate the moles of NH3 and O2

that is Pv= n RT

where; P= pressure,

V= volume,

n = number of moles,

R=gas constant = 0.0821 l .atm/ mol.K

make n the formula of the subject by diving both side by RT

n = PV /RT

The moles of NH3

n= (1.50 atm x 12.5 L) /( 0.0821 L. atm /mol.k x 298 K) =0.766 moles

The moles of O2

=(1.1 atm x 18.9 L) / ( 0.0821 L. atm/ mol.k x 323 K) = 0.784 moles

write the reaction between NH3 and O2

4 NH3 + 5 O2 →4 No +6H2O

from equation above 0.766 moles of NH3 reacted to produce

0.766 x 6/4 =1.149 moles of H2O

0.784 moles of O2 reacted to produce 0.784 x 6/5=0.9408 moles of H20

since O2 is totally consumed, O2 is the limiting reagent and therefore the moles of H2O produced= 0.9408 moles

mass of H2O = moles x molar mass

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mass = 18 g/mol x 0.9408 moles= 16.93 grams

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3 years ago

How many grams of O₂ are required to react completely with 14.6 g of Na to form sodium oxide, Na₂O?

Bad White [126]

The balanced chemical reaction is :

$O_2 + 4Na \ -> \ 2Na_2O$

Number of moles of Na, $n = \dfrac{14.6}{23} = 0.635 \ mol$ .

Now, from balance chemical reaction we can see that 1 mole of oxygen reacts with 4 moles of sodium.

So, number of moles of oxygen are :

$n = \dfrac{0.635}{4}\ mole$

So, amount of oxygen required is :

$m = \dfrac{0.635 \times 32}{4}\ gm\\\\m = 5.08 \ gm$

Therefore, 5.08 gram of oxygen will react with 14.6 gram of sodium.

7 0

3 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

docker41 [41]

Answer:

Explanation:

Recall that ΔH is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

5 0

3 years ago